[proofplan] (1) $\Leftrightarrow$ (2): the eigenspace dimension count from [Characterisations of Diagonalisability](/theorems/404) together with $\sum a_{\lambda_i} = n$ and $g_{\lambda_i} \leq a_{\lambda_i}$ forces equality iff diagonalisable. (1) $\Leftrightarrow$ (3): [Diagonalisability via Minimal Polynomial](/theorems/406) characterises diagonalisability by $M_\alpha$ having distinct linear factors, which is exactly $c_\lambda = 1$ for all eigenvalues. [/proofplan] [step:Show (1) $\Leftrightarrow$ (2) via the dimension count] Since $\chi_\alpha$ splits over $\mathbb{F}$, the distinct eigenvalues $\lambda_1, \dots, \lambda_k$ account for all roots: $\sum_{i=1}^k a_{\lambda_i} = n = \dim V$. By [Characterisations of Diagonalisability](/theorems/404), $\alpha$ is diagonalisable iff $\dim V = \sum_{i=1}^k g_{\lambda_i}$. By [Multiplicity Inequalities](/theorems/409), $g_{\lambda_i} \leq a_{\lambda_i}$ for each $i$. Since $\sum a_{\lambda_i} = n$: \begin{align*} \sum_{i=1}^k g_{\lambda_i} = n = \sum_{i=1}^k a_{\lambda_i} \quad \Longleftrightarrow \quad g_{\lambda_i} = a_{\lambda_i} \text{ for all } i. \end{align*} The forward implication holds because if two sums are equal and each summand on the left is at most the corresponding summand on the right, then each pair must be equal. [/step] [step:Show (1) $\Leftrightarrow$ (3) via the minimal polynomial characterisation] By [Diagonalisability via Minimal Polynomial](/theorems/406), $\alpha$ is diagonalisable iff $M_\alpha(t)$ is a product of distinct linear factors. Since $M_\alpha(t) \mid \chi_\alpha(t)$ by [Cayley-Hamilton](/theorems/407) and $\chi_\alpha$ splits, $M_\alpha$ also splits over $\mathbb{F}$. The condition "distinct linear factors" means exactly that each root of $M_\alpha$ has multiplicity $1$. The roots of $M_\alpha$ are precisely the eigenvalues of $\alpha$ (by [Multiplicity Inequalities](/theorems/409), $c_\lambda \geq 1$ iff $\lambda$ is an eigenvalue). Therefore "$M_\alpha$ has distinct linear factors" is equivalent to $c_\lambda = 1$ for every eigenvalue $\lambda$. [/step]